= Solution
\b[No.] The missing point is not assumed to complete $S$ as a smooth surface, so the hypotheses of the global <Gauss-Bonnet theorem> do not apply to $S\cup\{p\}$.
For a concrete counterexample, rotate a smooth simple arc $(r(s),z(s))$, parametrized by arc length for $0\leq s<L$, about the $z$-axis. Arrange that it starts smoothly on the axis,
$$
r(0)=0,\qquad r'(0)=1,
$$
stays in $r>0$ for $0<s<L$, and converges to another point of the axis as $s\uparrow L$, with
$$
r'(s)\longrightarrow-a,\qquad0\leq a<1.
$$
The endpoint is omitted. The resulting <surface of revolution> is diffeomorphic to $\mathbb S^2$ minus one point, and adjoining its limiting endpoint makes the set compact.
For an arc-length surface of revolution,
$$
K=-\frac{r''}{r},
\qquad
dA=r\,ds\,d\theta.
$$
Therefore the <total Gaussian curvature of a punctured surface with a singular compactification point> is
$$
\int_SK\,dA
=-2\pi\int_0^Lr''(s)\,ds
=2\pi(1+a),
$$
which can take values different from $4\pi$. Compactness of the one-point closure alone therefore does not determine the integral.
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